Question

Show
7that if a, b, c and d are integers, where! = 0, such that a | c and b | d, then ab | cd?

Answer 1

My interpretation:\(\\[0.6em]\)
\(\color{black}{{\rm Show{\small ~~}that{\small ~~}\forall}a,b,c,d\in\mathbb{Z},{\small~~}{\rm if}{\small~~}a|c{\small~~}{\rm and}{\small~~}b|d,{\small~~}{\rm then}{\small~~}ab|cd.}\)

Answer 2

Proof: \(\\[1.4em]\)
Since \(a|b\), therefore \(b=am\) for some \(m\in\mathbb{Z}\).
Since \(c|d\), therefore \(d=cn \) for some \(n\in\mathbb{Z}\).
Consequentially, \(cd=(am)(cn)=ac(mn)\), with \(mn\in\mathbb{Z}\).
Therefore \(ab|cd\).